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Author(s): D. COLLIER and I. D. MORRIS
Article Title: Approximating the maximum ergodic average via periodic
orbits
Year of publication: 2008
Link to published version:
Ergod. Th. & Dynam. Sys.(2008),28, 1081–1090 c 2008 Cambridge University Press
doi:10.1017/S014338570700082X Printed in the United Kingdom
Approximating the maximum ergodic average
via periodic orbits
D. COLLIER† and I. D. MORRIS‡
†School of Mathematics, University of Manchester, Manchester M13 9PL, UK (e-mail: [email protected])
‡Mathematics Institute, University of Warwick, Coventry CV4 7AL, UK (e-mail: [email protected])
(Received9February2006and accepted in revised form4September2007)
Abstract. Letσ:6A→6Abe a subshift of finite type, letMσ be the set of allσ-invariant Borel probability measures on6A, and let f:6A→Rbe a H¨older continuous observable.
There exists at least oneσ-invariant measureµwhich maximizesR f dµ. The following question was asked by B. R. Hunt, E. Ott and G. Yuan: how quickly can the maximum of the integralsR
f dµbe approximated by averages along periodic orbits of period less thanp? We give an example of a H¨older observable f for which this rate of approximation is slower than stretched-exponential in p.
1. Introduction
Letσ:6A→6A be a subshift of finite type, letMσ be the set of allσ-invariant Borel probability measures on6A, and let f:6A→Rbe H¨older continuous. There is at least
one invariant measureµ∈Mσ, which we term amaximizing measurefor f, such that
Z
f dµ=β(f):= sup
ν∈Mσ
Z
f dν.
In this paper we investigate the problem of finding invariant measures supported on periodic orbits which approximately realize this maximum. More specifically, if we let Mσ,pbe the the set of all ergodicσ-invariant measures supported on points fixed byσp, we investigate the quantity
Ef(p)= sup
ν∈Mσ
Z
f dν− sup
ν∈Mσ,p
Z
f dν,
introduced by Yuan and Hunt [11], which is the difference between the maximum integral β(f)of f and the ‘best approximating’ periodic orbit whose period divides p.
a specified period may be. These considerations motivate the work of Yuan, Hunt and Ott [7, 11].
Secondly, the quantityβ(f)has shown itself to be of interest in a number of situations in ergodic theory, both intrinsically [1, 2, 6, 8] and in application to existing problems [3, 4, 9]. It is therefore of interest to be able to compute this quantity accurately in numerical experiments. One obvious approach to this task would be to exhaustively compute ergodic averages of f along periodic orbits of length up to n, and take the supremum of these averages as an approximation toβ(f). The error incurred in this approximation would therefore equal inf1≤p≤nEp(f).
It is a classical result [10] thatS∞
p=1Mσ,pis dense inMσ in the weak-* topology, and soEf(p)→0 as p→ ∞for all continuous f. We wish to investigate the rate at which this convergence occurs when f is H¨older.
The behaviour ofEf(p)as p→ ∞is at present poorly understood. On the strength of numerical experiments [7] combined with rigorous analysis, Yuan and Hunt [11] observed that the sequence Ef(p)often decays to zero at an exponential rate, but were unable to prove this in generality. They then asked whether it could be shown thatEf(p)always decays exponentially.
The purpose of this paper is to answer this question in the negative. We have the following theorem.
THEOREM1. Let σ: 6A→6A be a subshift of finite type. There exists H¨older continuous f:6A→Rsuch that Ef(p)tends to0at a slower than stretched-exponential rate:logEf(p)=o(pε)for everyε >0.
We note that a result similar to Theorem 1 has recently been proved by Bressaud and Quas in [5], in which the quantity inf1≤p≤nEp(f)is considered. Bressaud and Quas were able to obtain upper and lower bounds that are superior to those in the present article, but with the weakness that their lower bound applies only along subsequences of integersn.
2. Notation and definitions
Let A be an irreducible aperiodicN×N matrix of zeros and ones. We define theshift spaceassociated withAto be the set
6A:= {x=(xi)i≥1:xi ∈ {1, . . . ,N}andA(xi,xi+1)=1 for alli≥1},
and define theshift mapσ:6A→6Aby
(σx)i =xi+1
for alli≥1. Givenθ∈(0,1), we define theθ-metricdθ on6Aby dθ (xi)i≥1, (yi)i≥1=θinf{n≥1:xn6=yn}.
We say that a function f:6A→Risθ-H¨older continuousif it is Lipschitz continuous with respect to the metricdθ. We fixθ∈(0,1)for the remainder of this paper.
We define a finite word to be a finite sequence ω=(ωi)n
i=1 taking values in the
set {1,2, . . . ,N}. We say that ω=(ωi)n
i=1 is compatible with the matrix A if A(ωi, ωi+1)=1 for alli<n. We define thelengthof the word(ω)ni=1to ben. We will
Approximating the maximum ergodic average via periodic orbits 1083
equal lengthnarerotation equivalentif there exists a non-negative integerr<nsuch that ω1
i =ω 2
i+r whenever 1≤i≤n−r, andω 1 i =ω
2
i+r−nwhenevern−r<i≤n. Whenω 1
and ω2 are rotation equivalent we write ω1'ω2. Words of distinct lengths are never rotation equivalent.
Given two words ω1 and ω2 with lengths n1 and n2, respectively, we define their concatenationω1·ω2 to be the word of lengthn1+n2 given by[ω1·ω2]i=ω1i when 1≤i≤n1, and[ω1·ω2]i =ω2
i whenn1+1≤i≤n1+n2. Given a finite list of words ω1, . . . , ωm, we denote the compound concatenation ω1·ω2· · · · ·ωm by Qm
k=1ωk.
Concatenation is associative. If ω1, ω2are finite words with lengths n1,n2 compatible
withA, their concatenation is compatible withAif and only ifA(ωn1 1, ω
2 1)=1.
For each p>0, we let p be the set of all words of length p which are compatible with A. We letp be the set of all wordsω∈p such that A(ωp, ω1)=1. Note that
ω∈pif and only if bothω∈pandω·ω∈2p. If x∈6A, it is clear thatσpx=x if and only if x
i=xi+p for alli≥1, if and only if there existsω∈p such thatxi+kp=ωi for all 1≤i≤p andk≥0. In this case we writex=π(ω). This defines a relationship between the sets Fixp= {x∈6A: σpx=x} andp, which is readily seen to be bijective. Moreover, we haveπ(ω1)=σjπ(ω2)for some j≥0 if and only ifω1'ω2.
Ifa=(ai)in=1is a finite word andbis either a finite word or an element of6A, we write a≺bif there isk≥0 such thatai =bi+k for all 1≤i≤n. In this case we say thatais a subwordofb.
3. Proof of Theorem 1
We begin with the following.
PROPOSITION3.1. Let K=σK be a closed non-empty subset of6A, and define fK(x)= −dθ(K,x)for all x∈6A. Clearly fKisθ-H¨older continuous. For eachω∈pwe define
ξ(ω,K):= inf
ω0'ω
{` >0: ∃a∈`such that a≺ω0and∀x∈K, a⊀x}.
Let p>0, and suppose thatsupω∈pξ(ω,K)≤p. Then
EfK(p)≥
1
pθ
supω∈pξ(ω,K)= 1 p ωinf∈p
θξ(ω,K).
Proof. Letp>0 andω∈p, whereξ(ω,K)≤p. We have logθ sup
ω0'ω
d(π(ω0),K)=logθ sup
ω0'ω
inf
x∈K dθ(π(ω 0),x)
= inf
ω0'ωsup x∈K
inf{` >0: x`6=π(ω0)`} = inf
ω0'ωsup{` >0: ∃x∈K such thatxi=π(ω
0)
i∀1≤i< `} = inf
ω0'ω{` >0: ∃a∈`such thatai=π(ω
0)
i ∀1≤i≤`and∀x∈K, a⊀x}
= inf
ω0'ω
{` >0: ∃a∈`such thata≺π(ω0)anda⊀x∀x∈K} = inf
ω0'ω
{` >0: ∃a∈`such thata≺ω0anda⊀x∀x∈K}
where we have usedξ(ω,K)≤p in the second-from-last equality. Since K is closed, σ-invariant and non-empty, the Krylov–Bogolioubov theorem shows that there exists µ∈Mσ such thatµ(K)=1. It follows thatβ(f)=0, and so
EfK(p)=σpinfx=x
1
p p−1
X
j=0
d(σjx,K)
≥ 1
p σpinfx=x sup 0≤j<p
d(σjx,K)
= 1 p ωinf∈p
sup
ω0'ω
d(π(ω0),K)
≥ 1 p ωinf∈p
θξ(ω,K),
as required. 2
To prove Theorem 1, it therefore suffices to construct a non-empty compact set
K=σK⊆6A such that supω∈
p ξ(ω,K)=o(p
ε)as p→ ∞. The remainder of this
section is dedicated to this task. We will construct the invariant setK recursively, using a sequence of sets of wordsMnof increasing length. The setK will then arise as a limit of these sets.
Let M1 be a subset of `1 for some positive integer`1 such thatm1=CardM1 is
divisible by 216. We require thatM1have the following properties.
Definition 3.2. There exists an integerP≥1 and a wordz=(zi)iP=1∈Psuch that: (i) everyω∈M1satisfieszi =ωi for all 1≤i≤P;
(ii) if ω=ω1·ω2 where ω1, ω2∈M1, and z
i=ωi+r for all 1≤i≤P, then either r=`1orr=0;
(iii) A(ω1` 1, ω
2
1)=1 for all ω
1, ω2∈M
1; that is, ω1·ω2∈2`1 for every pair ω 1,
ω2∈M 1.
The reader may verify that such a setM1can be constructed for any prescribedm1and matrixA.
A sequence of setsMn with cardinalitiesmnconsisting of words of length`n will be defined in an inductive fashion, starting with the setM1. We begin by introducing some integer sequences which will be crucial to our construction.
Definition 3.3. Given an integerm1divisible by 216, letq1=t1=m1/4. Define sequences
(mn)n≥1, (qn)n≥1 and (tn)n≥1 as follows. Given the integer tn, let mn+1=4tn and qn+1=(1/4)mn+1=4tn−1, and letτn+1be the unique positive real number such that
τ √
τn+1 n+1 =4
tn−1=q n+1.
Then define
tn+1=4b(tn/
√
τn+1)c−1.
Approximating the maximum ergodic average via periodic orbits 1085
LEMMA3.4. For each n≥1, we have tn|qn, mn/qn=4 and tn≥214. Moreover, the sequence(tn)n≥1satisfies tn/
√
tn+1≤16tn1/4for all n≥1and lim
n→∞
tn tn+1
=0.
Proof. The first two statements are clear. We consider the sequence(tn)n≥1; the definition
implies that
1
16τn+1=4( tn−1/
√
τn+1)−2≤4b(tn/
√
τn+1)c−1=t
n+1≤4(tn/
√
τn+1)−1≤τ n+1,
for everyn≥1. We proceed inductively. Given thattn≥214, notice that √
τn+1logτn+1=(tn−1)log 4. We thus have
tn= √
τn+1log4τn+1+1> √
τn+1,
and therefore
log4τn+1<2 log4tn.
We deduce
√ tn+1≥
1 4
√
τn+1=
tn−1 4 log4τn+1
≥ tn−1
8 log4tn ≥ 1
8t
3/4
n −
1 8,
where we have used the elementary inequality log4t≤t1/4for allt≥16. Thus
tn+1≥641(tn3/4−1)2≥2−6(221−211 √
2+1) >214, for everyn≥1, making it clear thattn/
√
tn+1≤16tn1/4. One may easily use the above to show thattn+1≥tn+1/64 for eachn≥2, which implies thattn→ ∞; since for eachn we have
0≤ tn tn+1
≤ 4tn tn3/2−2t
3/4
n +1
,
it follows that limn→∞tn/tn+1=0 as required. 2
For an integer n≥1, a finite word a=(ai)mi=1 and a finite or infinite word b,
we shall write a≺nb if there is k≥0 such that ai=bi+k`n for all 1≤i≤m. The
distinction between≺and≺n will be important since we will construct wordsω∈Mn+1
as concatenations of wordsa,b,c, . . .∈Mn. For example, ifa,b,c∈Mn, then it is true thatb·c≺na·b·c·aandc·b⊀na·b·c·a; however, the statementc·b≺a·b·c·a could be either true or false, depending on the subword structure of the wordsa,bandc.
Given n≥1 and the set Mn, we construct the set Mn+1 as follows. Recalling that mn=4qn, partition Mn intoqn disjoint setsCnk of cardinality 4, where 1≤k≤qn. For eachk, we will writeCnk= {ckj:1≤ j≤4}. Define
In=(i1,i2, . . . ,iqn)∈ {1,2,3,4} qn:i
m=im+tn for all 1≤m≤qn−tn .
For eachqn-tuple(i1, . . . ,itn)∈In, we construct the word
ω(i1,...,iqn):= qn
Y
k=1 cik
LetMn+1= {ω(i1,...,iqn):(i1, . . . ,iqn)∈In}, then clearly
CardMn+1=CardIn=4tn =mn+1,
in accordance with Definition 3.4. We remark that Definition 3.2(iii) implies that
Mn⊆`n for everyn≥1. The key features of the above construction are summarized
in the following lemma.
LEMMA3.5. The following are direct consequences of the definition of the sets Mn. (i) IfQqn
k=1ckjk∈Mn, where n≥2and each c k jk ∈C
k
n−1, then ckjk∈C 1
n−1if and only if k=1.
(ii) Ifω1=ckj11·ckj22 with ckj11∈Cnk1and ckj22 ∈Cnk2, then there exists N>n andω2∈MN such thatω1≺nω2if and only if k2=1+k1modqn.
(iii) If ω1=Qk0
+tn k=k0 c
k
jk, where each c k
j∈C k
n and k0≤qn−tn with jk06= jk0+tn, then
ω1⊀nω2for allω2∈MN whenever N≥n+1.
(iv) If ω1≺nω2∈Mn+1, ω1≺nω3∈Mn+1 and ω1 has length at least tn`n, then ω2=ω3.
(v) For each n≥1we have
`n+1=qn`n=14mn`n and hence
`n+1=`1 n
Y
k=1 qk.
The proof is clear. 2
The following lemma allows us to pass from the relation≺to the relation≺n, and thus make use of Lemma 3.5.
LEMMA3.6. Let a=(ai)i`=n1∈Mn and ω=(ωi)
`N
i=1∈MN where n<N . Suppose that there is r≥0such that ai=ωr+i for all1≤i≤`n. Then`n|r .
Proof. We first prove the case n=1. Let ω=Qm
k=1ωk where each ωk∈M1, and
supposeai=ωr+i for all 1≤i≤`1. There existsk∗such thatk∗`1≤r+i< (k∗+2)`1
for all 1≤i≤`1, so that if we let ωˆ =ωk∗·ωk∗+1, then zi =ai= ˆωi+r−k∗`
1 for all
1≤i≤P≤`1 by Definition 3.2(i). By Definition 3.2(ii) we have eitherr−k∗`1=`1
orr−k∗`1=0, and so`1|ras required.
We proceed by induction onn. Leta=Qqn−1
k=1 aˆk andω=
Qm
k=1ωˆk, with each aˆk, ˆ
ωk∈M
n−1. Letaˆ1=(aˆi1)`i=n−11. Since aˆ 1
i =ωr+i for all 1≤i≤`n−1, we have`n−1|r by the induction hypothesis. Since each ωˆk has length `n−1, it follows that there is s=r/`n−1>0 such thatQ
qn−1 k=1 aˆk=
Qk=s+qn−1
k=s ωˆk. By Lemma 3.5(i) we haveaˆk∈Cn1−1
if and only if k=1, and similarly ωˆk∈Cn1−1 if and only if k≡1 modqn. Since ˆ
ωs+1∈C1
n−1 it follows that s≡0 mod qn. Since r=s`n−1 and `n=qn−1`n−1, we
deduce that`n|r. 2
Approximating the maximum ergodic average via periodic orbits 1087
COROLLARY3.7. Letω=(ωi)p
i=1be admissible. Let a,b∈Mn and write a=(ai)`i=n1, b=(bi)i`=n1. If there are r,s>0such that ai=ωi+r, bi =ωi+s for all1≤i≤`n, then
`n|r−s.
We may now prove the next lemma.
LEMMA3.8. The following constraints on admissibility hold.
(i) If ω=ck1 j1 ·c
k2 j2 with c
k1 j1 ∈C
k1
n and ckj22∈Cnk2, then ω is admissible if and only if k2=1+k1modqn.
(ii) Ifω=Qk0+tn k=k0 c
k
jk with each c k
jk∈C k
nand jk0 6=jk0+tn where k0≤qn−tn, thenωis not admissible.
Proof. Apply Lemma 3.5 and Lemma 3.6. 2
LEMMA3.9. Let a,b∈Mnwhere n≥2, and suppose that the word a·b is inadmissible. Then a·b has an inadmissible subword of length less than or equal to2tn−1`n−1. Proof. Let a=Qqn−1
k=1 ak and b=
Qqn−1
k=1 bk where each ak,bk∈Mn−1. Define u=
Qqn−1
k=qn−1−tn−1+1a
k andv=Qtn−1
k=1bk, each being an admissible word of lengthtn−1`n−1.
Clearly u·v≺a·b. Suppose, for a contradiction, that u·v≺ω1·ω2=ω, say, whereω1, ω2∈Mn andωis admissible. Let (u·v)i=ωi+r for all 1≤i≤2`n−1. By
Lemma 3.6,r=`n−1sfor somes. Suppose thats≥qn; thenv≺nω2and soω2=bby
Lemma 3.5(iv). Thuss=qn and henceu≺nω1andω1=a. Thereforeω=a·band so
is not admissible. It follows thatu·v is inadmissible; this word has length 2tn−1`n−1as
required. 2
LEMMA3.10. Letωbe a word of length`≥`n, where n≥2. Then eitherω'Qm k=1ωk with eachωk∈Cnk−1, or there existsω0'ωwhich has an inadmissible subword of length less than or equal to3`n−1.
Proof. Suppose that for allω0'ω, every subword ofω0of length 3`n−1is admissible. Let
`=s`n−1+r with 0≤r< `n−1. We claim that there existωk∈Mn−1and a word ω∗
of lengthr such thatω'Qs
k=1ωk·ω
∗. We will show that(ω i)
m`n−1 i=1 '
Qm
k=1ωk for all m≤sby induction onm.
Clearly, any admissible word of length 3`n−1 must include some a∈Mn−1 as a
subword, soωmust include such a subword. Taking a rotation equivalent ofωif necessary, we deduce that there existsω1∈Mn−1such thatωi=ω1i for all 1≤i≤`n−1. This proves
the casem=1.
Given that(ωi)mi=`1n−1=
Qm
k=1ωkwith eachωk∈Mn−1, consider the wordb=(bi) 2`n−1 i=1
defined by bi=ω(m−1)`n−1+i for all 1≤i≤2`n−1, which is well-defined as long as m+1≤s. Sinceb has length 2`n−1<3`n−1, it is admissible and so there exist N, M, t>0 and a=QM
k=1ak∈MN with eachak∈Mn−1 such that bi=at+i for all 1≤i≤ 2`n−1. By Lemma 3.6 we have t=`n−1t for some t, so thatb=at+1·at+2. Thus
(ωi)(m +1)`n−1 i=m`n−1+1=(bi)
2`n−1 i=`n−1+1=a
t∈M
n−1 as required to prove the case m+1. This
Since for anyω0'ω, every subword ofω0 with length less than or equal to 3`n−1 is
admissible, the wordz=ωs·ω∗·ω1must be admissible. By Corollary 3.7 this implies
`n−1|rand hencer=0, thus completing the proof. 2
LEMMA3.11. Letω=Qm
k=1ωk for someω1, . . . , ωm∈Mn with m≥1. Then at least one of the following holds:
(a) there existsω0'ωsuch thatω0has an inadmissible subword with length less than or equal to2`n−1tn−1;
(b) there existsω0'ωsuch thatω0has an inadmissible subword with length less than or equal to(1+tn)`n, and m≥qn; or
(c) there existmˆ >0andωˆ1, . . . ,ωˆmˆ ∈Mn+1such that m=`n+1m andˆ ω'Qmkˆ=1ωˆk. Proof. Suppose that case (a) does not hold. Writingω=Qm
i=1c ki
ji with everyc ki ji ∈C
ki n, this assumption implies, via Lemma 3.8(i) and Lemma 3.9, that for everyi<mwe have
ki1=1+ki modqn. Clearly, there exists a rotation equivalent of ωwhich includes the
wordckm jm ·c
k1
j1. Again by Lemma 3.8(i) and Lemma 3.9, our assumption that case (a) does
not hold implies that this word is admissible, and hencek1=km+1 modqn. It follows that the sequenceki must take every value in the range 1, . . . ,qn an equal number of times. Taking a rotation equivalent if necessary, we have
ω= S
Y
s=1 qn
Y
k=1 ckj
k,s
for some integerS≥1 and some sequencejk,s. Note in particular thatm≥qn(as required for (b) to hold).
We now suppose also that (b) does not hold. To show thatω is a concatenation of elements ofMn+1, and hence that (c) holds, it remains to show that jk,s=jk+tn,s for all
1≤s≤S and 1≤k≤qn−tn. If this is not the case, thenωmust include a subword of the form
k0+tn
Y
k=k0 ckj,s
for some fixedssuch that jk0,s 6=jk0+tn,s, which is inadmissible by Lemma 3.8(ii) and has
length(1+tn)`n, implying the case (b). 2
Combining the above lemmata, we obtain the following.
PROPOSITION3.12. Define a set K⊆6Aby letting x∈K if and only if every subword of x is admissible. Then K is closed, satisfiesσK =K , and is non-empty. Moreover,
sup
ω∈p
ξ(ω,K)=o(pε)
for everyε >0.
Proof. ThatK is a non-empty subset of6Afollows from the fact thatMn⊆`n for every n≥1. The proof that K is closed and σK=K is straightforward. Given ω∈p and `n≤p< `n+1, we will attempt to bound the quantity
ξ(ω,K)= inf
ω0'ω{` >0: ∃a∈`such thata≺ω
0
Approximating the maximum ergodic average via periodic orbits 1089
Suppose thatξ(ω,K) >3`n−1. Then by Lemma 3.10 we haveω'Qmi=1c ki
ji, where each cki
ji ∈C ki
n−1andm=p/`n−1. Lemma 3.11 then implies that eitherξ(ω,K)≤2`n−2tn−2,
orξ(ω,K)≤`n−1(1+tn−1), orω'Qimˆ=1c ki
ji where eachc ki ji ∈C
ki
n andmˆ =p/`n. In the last of these three alternatives, apply Lemma 3.11 again to see that, since p< `n+1, either
ξ(ω,K)≤2`n−1tn−1, or p≥`nqn andξ(ω,K)≤`n(1+tn)≤2ptn/qn. We conclude that, in any case,
ξ(ω,K)≤max
3`n−1,2`n−2tn−2, `n−1(tn−1+1),2`n−1tn−1,
2ptn
qn
=max
2`n−1tn−1,
2ptn
qn
.
Hence, for everyn≥3 andε >0,
sup
`n≤p<`n+1
sup
ω∈p
1
pεξ(ω,K)≤max
2`n−1tn−1`−nε,2 `1−ε
n+1tn qn
≤2 max`1−ε n−1tn−1q
−ε n−1, `
1−ε
n tnqn−ε =2`1 −ε n tnqn−ε.
To complete the proof, therefore, it suffices to show that tnqn−ε`n→0 for every ε >0. Using Lemma 3.5(vii), Lemma 3.4 and Definition 3.3, we have
log4(tnqn−ε`n)≤log4 4(tn−1/
√
τn)−εtn−1q1` 1
n−1
Y
k=2 qk
!
≤log4(q1`1)+ tn−1 √
tn
−εtn−1+ n−2
X
k=1 tk
= −εtn−1+ n−2
X
k=1
tk+O(tn1−/41).
Elementary analysis then shows that since tn−1/tn tends to zero, Pn −2
k=2tk=o(tn−1)
and so
log4(tnqn−ε`n)→ −∞,
as required. 2
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